A trader marked the selling price of an article at 10% above the cost price. At the time of selling, he allows certain discount and suffers a loss of 1%. He allowed a discount of :
A. 10 %
B. 10.5 %
C. 11 %
D. 12.5 %
A. 10 %
B. 10.5 %
C. 11 %
D. 12.5 %
A. 8
B. 12
C. 6
D. 14
E. None of these
Explanation:
Let the numbers be x, x + 2, x + 4, x + 6, x + 8 and x + 10.
Given (x + 2) + (x + 10) = 24
=> 2x + 12 = 24 => x = 6
The fourth number = x + 6 = 6 + 6 = 12.
A. no profit, no loss
B. 20% profit
C. less than 20% profit
D. more than 20% profit
Explanation:
Let the article be worth Rs. x.
C.P. 90% of Rs. x = Rs. 9x/10
S.P. = 110% of Rs. x = Rs. 11x/10
Gain = (11x/10 – 9x/10) = Rs. x/5
Gain % = x/5 * 10/9x * 100 = 22 2/9 % > 20%
A. 1:3
B. 3:1
C. 1:2
D. 2:1
A. 5days
B. 6days
C. 7days
D. 8days
Explanation:
Let us assume that Arun uses X units of petrol everyday.
So the amount of petrol in the tank when it is fuel will be 10X.
If he started using 25% more petrol every day, then the amount of petrol he how uses everyday will be
X (1 +25/100) =1.25x
Therefore, number of days his petrol will how last = Amount of petrol in tank / amount of petrol used everyday = 10x/1.25x = 10/1.25 = 8 Days
A. 4500 M3
B. 27000 M3
C. 3000 M3
D. 2700 M3
(3000 * 2 * 5)/60 = 4500 M3
A. 8
B. 12
C. 6
D. 2
Let the sides of the rectangle be l and b respectively.
From the given data,
(√l2 + b2) = (1 + 108 1/3 %)lb
=> l2 + b2 = (1 + 325/3 * 1/100)lb
= (1 + 13/12)lb
= 25/12 lb
=> (l2 + b2)/lb = 25/12
12(l2 + b2) = 25lb
Adding 24lb on both sides
12l2 + 12b2 + 24lb = 49lb
12(l2 + b2 + 2lb) = 49lb
but 2(l + b) = 28 => l + b = 14
12(l + b)2 = 49lb
=> 12(14)2 = 49lb
=> lb = 48
Since l + b = 14, l = 8 and b = 6
l – b = 8 – 6 = 2m.